Natural gas is primarily methane, CH 4 , and is used to heat homes. A typical ho
ID: 890214 • Letter: N
Question
Natural gas is primarily methane, CH4, and is used to heat homes. A typical home is approximately 2000 ft2 and the ceilings are 8 ft high. The following data may be helpful: The heat capacity of air is 1.01 J/g·K and the and the enthalpy of combustion of methane is -890.8 kJ/mol. Assume that the molecular weight of air is the same as nitrogen, its major component.
How many grams of methane are required to raise the temperature in the home from 39°F to 72°F?
_______g
How many grams of CO2 does this reaction produce?
_______g
Explanation / Answer
volume of home = 2000*8 = 16000 ft^3 = 453069.55 L
heat capacity of air = 1.01 J/g·K
enthalpy of combustion of methane DHcom = -890.8 kJ/mol
molecular weight of air = nitrogen = 28 g/mol
mass of air present in the home w = PV*mWt/RT
p = atmospheric pressure = 1 atm
T = 39 F = 277.039 k
mwt = 28 g/mol
v = 453069.55 L
R = 0.0821 L.atm.k-1.mol-1
w = (1*453069.55*28)/(0.0821*277.039)
= 557749.0488 grams
heat absorbed q = m*s*DT
m = mass of air = 557749.0488 grams
s= heat capacity of air = 1.01 J/g·K
DT = 72-39 = 33
q = 557749.0488* 1.01*33
heat absorbed q = 18589.8 kj
1 mole of CH4 gives = 890.8 kJ/mol
No of moles of CH4 required = 18589.8/890.8
= 20.9 mole
CH4(g)+ 2O2 -----> CO2(g) + 2H2O(g)
mass of methane required = 20.9*16 = 334.4 grams
from equation
1 mole CH4 = 1 MOLE CO2
Mass of CO2 PRODUCED = 20.9*44 = 919.6 GRAMS
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